# Convergence in the Generalized Fibonacci Sequence to the golden ratio

Many posts have shown the convergence of the ratio of two consecutive terms in the Fibonacci sequence $$F_{n+1} = F_n + F_{n-1}$$ when the starting values are $$F_0 = 0$$ and $$F_1 = 1$$. How do we show that this holds for any starting values that are integers $$a$$ and $$b$$ (at least one is nonzero). That is, for large $$n$$:

$$\lim_{n \to \infty} \dfrac{F_{n+1}}{F_n} =\dfrac{1+\sqrt{5}}{2} = \Phi$$

Does that hold for any $$a,b \in \mathbb{R}$$ provided not both are zero?

## 1 Answer

When you solve the recurrence $$a_n=a_{n-1}+a_{n-2}$$, you find that the solutions all have the form

$$a_n=A\varphi^n+B\hat\varphi^n\,,$$

where $$\varphi=\frac12(1+\sqrt5)$$ and $$\hat\varphi=\frac12(1-\sqrt5)$$; the values of $$A$$ and $$B$$ are determined by the initial values $$a_0$$ and $$a_1$$. And $$\varphi\hat\varphi=-1$$, so,

\begin{align*} \frac{a_{n+1}}{a_n}&=\frac{A\varphi^{n+1}+B\hat\varphi^{n+1}}{A\varphi^n+B\hat\varphi^n}\\\\ &=\frac{\varphi+\frac{B}A(-1)^n\hat\varphi^{2n+1}}{1+\frac{B}A(-1)^n\hat\varphi^{2n}}\,, \end{align*}

and

$$\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=\lim_{n\to\infty}\frac{\varphi+\frac{B}A(-1)^n\hat\varphi^{2n+1}}{1+\frac{B}A(-1)^n\hat\varphi^{2n}}=\varphi$$

as long as $$A\ne0$$. If $$A=0$$, however,

$$\frac{a_{n+1}}{a_n}=\hat\varphi=-\frac1\varphi\,.$$